[EM] Random and reproductible tie-breaks
Stéphane Rouillon
stephane.rouillon at sympatico.ca
Thu Sep 25 06:54:51 PDT 2008
Hello Raph,
maybe I was not clear. I mean an equiprobable but determined tie-breaker.
First if we want to break a tie. Using your data,
>A wins: 1234
>B wins: 6542
>C wins: 2539
B wins and no need for any tie-breaker.
However to determine a winner in case for triple equality
I would proceed like that:
1) establish an index for winner using alphabetical order (A is 1, B is 2
and C is 0)
2) divide the tie value by the number of equally concerned candidates
For
>A : 1234
>B : 1234
>C : 1234
we obtain 1234 = 411 x 3 +1. Thus A wins.
For
>A : 6542
>B : 6542
>C : 6542
we obtain 6542 = 2180 x 3 +2. Thus B wins.
For
>A : 2539
>B : 2539
>C : 2539
we obtain 2539 = 846 x 3 +1. Thus A wins.
Using total turnout is less random and easier to manipulate for fraud during
a recount.
Tie-breakers have no need to be clone-independent. You cannot predict a tie
will occur
before the election and it would be ridiculous to strategize on this
hypothesis.
>From: "Raph Frank" <raphfrk at gmail.com>
>To: "Stéphane Rouillon" <stephane.rouillon at sympatico.ca>
>CC: easmith at beatrice.rutgers.edu, election-methods at electorama.com
>Subject: Re: [EM] Random and reproductible tie-breaks
>Date: Thu, 25 Sep 2008 11:25:24 +0100
>
>On 9/25/08, Stéphane Rouillon <stephane.rouillon at sympatico.ca> wrote:
> > Hello Allen,
> >
> > simply using the number of ballots involved in the tie is enough.
>Compare
> > its rest using euclidian divison by the number of involved candidates to
>the
> > alphabetical rank of the candidates.
>
>Just to clarify, you mean divide the number of ballots by the number
>of candidates involved in the tie and use the remainder (rest?) to
>decide who wins (candidates sorted alphabetically)?
>
> > Simple, effective and greatly equiprobable. It works for winner
>selection
> > as for elemination rounds.
>
>However, it isn't clone independent. Random ballot is used to achieve
>that.
>
>What about reversing the digits in the number of ballots and using
>that as an index.
>
>For example, assuming a tie between A, B and C
>
>Check each ballot to see who would win if that ballot is picked
>
>A wins: 1234
>B wins: 6542
>C wins: 2539
>Turnout: 10315
>
>Reverse turnout: 51301
>Max possible: 99999 (with 5 digits)
>
>Rank used is ( Reverse / Max possible) * ( turnout - 1 )
>
>= 51301/99999 * (10315 - 1) = 5291 (rounded to nearest)
>
>A wins for 0 - 1233
>B wins for 1234 - 7775
>C wins for 7776 - 10314
>
>Thus in this case B wins. If the turnout had been 10319, then C would
>have won as the rank would have been 9417.
>
>It might require some tweaking to make it so that it gets the
>probabilities exactly right.
>
>Ofc, the process assumes that the number of ballots is reasonably
>solid. I am not sure that a tie could occur in a real election that
>was solid over multiple recounts. I think it would come down to which
>ballots the courts includes/excluded rather than the official tie
>breaking rule.
>
>If the tie breaking rule needs multiple rounds, the turnout could be
>squared and the result used for the 2nd round. Aternatively, a
>different base could be used to express the numbers.
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